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Homework Help: Mixing Problem HELP!

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A tank contains `100` kg of salt and `2000` L of water. A solution of a concentration `0.025` kg of salt per liter enters a tank at the rate `6` L/min. The solution is mixed and drains from the tank at the same rate.

    Find the equation for the amount of salt in the tank after t hours.

    3. The attempt at a solution

    I have 0.15kg/min as my rate in and y(t)/333.33 as my rate out.

    Which I put together in the form

    dy/dt = (5 - y(t))/(333.33)

    Split it into

    int of (dy/5 - y) = int of (dt/333.33)

    -ln(5 - y) = t/333.33 + C

    y(0) = 100, so C = -ln(-95)

    -ln(5 - y) = t/(333.33) - ln(-95)

    5 - y = -95e^(-t/333.33)

    y(t) = 5 + 95e^(-t/333.33)

    But that's wrong. I know I've asked alot of stuff today, but this one is driving me crazy, I've tried everything I could think of and nothing came out right. Help!
     
  2. jcsd
  3. Mar 17, 2008 #2
    This involves chemistry friend.

    n1v1=n2v2
     
  4. Mar 17, 2008 #3
    I'm...fairly sure it doesn't, because we've done problems like this before in Calculus without any Chemistry.
     
  5. Mar 17, 2008 #4

    dynamicsolo

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    Homework Helper

    This is correct. I believe you dropped a decimal place, because the next line should be

    dy/dt = (50 - y(t))/(333.33) ,

    which now gives you the incoming mass rate of 0.15 kg/min. The asymptotic mass of salt in the tank should be 50 kg
    ( = 0.025 kg/L ยท 2000 L), which is what the corrected mass function

    y(t) = 50 + 50e^(-t/333.33)

    will give you.

    I use decimals, rather than fractions, which avoided this headache...
     
  6. Mar 17, 2008 #5
    I never said that it cannot be solved without using chemistry.
    I wanted to say that this question becomes a piece of meat when solved using chemistry.
     
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